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27-36 Antibiotic-resistant bacteria have an enzyme, penicillase, that catalyzes the composition of the antibiotic. The molecular mass of penicillase is 30,000g/mol. The turnover number of the enzyme at 28°C is 2,000 s-1. If 6.4μg of penicillase catalyzes the destruction of 3.11mg of amoxicillin, an antibiotic with a molecular mass of 364 g/mol, in 20s at 28°C, how many active sites does the enzyme have? From the lecture notes on Monday, we know that the turnover or catalytic constant kcat is equal to Vmax/[E]. From the data given above, we can determine the velocity at which amoxicillin is being catalyzed. This is done by changing the amount of amoxicillin catalyzed into moles and dividing by time(in this case it will be 20s). Calculation: 3.11e 6 gamoxicillin 6.4e 6 gpenicilla se v 1mol 8.54e 6molamoxicillin 364 g 1mol 2.13e 10molespenicillase 30,000 g moles 8.54e 6moles amoxicilli n 4.27e 7mol / s time 20s Since there is an excess amount of substrate we can say that the v obtained above is equal to our Vmax. So the following calculation can be done. Calculation: Vmax v Vmax [E] V 4.27e 7 mol / s [ E ] max 2.135e 10mol penicillas e k cat 2000s 1 k cat Since the description of the turnover number is maximum rate per mole of enzyme active sites one can say that penicillase only has one active site, which is figured out by taking the following ratio: [E] moles needed to catalyze certain amount of amoxicilli n 2.135e 10molpenicil lase 1.00activesites 2.13e 10molpenicil lase # active sites